Mastering Exponential Variables: How to Solve e^x + e^-x = 4 with Ease

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To solve the equation e^x + e^-x = 4, substituting u for e^x simplifies the problem to u + 1/u = 4. This leads to the quadratic equation u^2 - 4u + 1 = 0 after multiplying both sides by u. The roots of this equation can be found using the quadratic formula, yielding solutions for u. The final step involves taking the natural logarithm to find x from the values of u. This approach clarifies the solution process and resolves the initial confusion.
Chiborino
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My friend just came to me with this problem:
e^x + e^-x = 4
I aced precalc, and I still have no idea where to start on this one. I've factored, natural logged, factored and then natural logged... and I keep running into dead ends.

My graphing calculator puts the answer at +/-1.3169 or something like that.

It's not at all urgent, I'm just curious since I feel that I should know how to solve this problem.
 
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If you put u=e^x then e^(-x)=1/u. So it becomes quadratic equation in u. Is that enough of a hint?
 
I'm not entirely sure I see how u+1/u can be plugged into the quadratic formula.

Could you please elaborate?
 
u+1/u=4. Multiply both sides by u.
 
Oh... I feel kind of dumb now. >_>
Thanks much.
 

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